I am reading "Topology 2nd Edition" by James R. Munkres.
The following exercise is in this book:
P.35
Exercise 10.
Show that every positive $a$ has exactly one positive square root, as follows:(a) Show that if $x>0$ and $0\leq h<1$, then $$(x+h)^2\leq x^2+h(2x+1),$$$$(x-h)^2\geq x^2-h(2x).$$
(b) Let $x>0$. Show that if $x^2<a$, then $(x+h)^2<a$ for some $h>0$; and if $x^2>a$, then $(x-h)^2>a$ for some $h>0$.
(c) Given $a>0$, let $B$ be the set of all real numbers $x$ such that $x^2<a$. Show that $B$ is bounded above and contains at least one positive number. Let $b=\sup B$; show that $b^2 = a$.
(d) Show that if $b$ and $c$ are positive and $b^2=c^2$, then $b=c$.
When we solve (b), we can use the result of (a). I think this is the author's intention.
But, if $x^2>a$, then it is obvious that $(x-h)^2>a$ for very large $h>0$. For example let $h:=x+a+1$.
I think the author intended that "if $x^2>a$, then there exists $h>0$ such that $(x-h)^2>a$ and $x-h>\sqrt{a}$".
But we have not yet shown that $\sqrt{a}$ exists.
How should I modify "if $x^2>a$, then $(x-h)^2>a$ for some $h>0$"?
The important part is that the squared number remains positive, otherwise you are looking for solutions of $x^2 = a$, rather than verifying that $f(y) = \sqrt{y}$ is well-defined for positive numbers. You could add the condition that $x-h > 0$.